洛谷P3832 [NOI2017]蚯蚓排队【链表 + 字符串hash】

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/Mychael/p/8996326.html

本文提供洛谷P3832题目的详细解答过程，采用字符串哈希结合哈希表的方法进行查询，并通过暴力维护1、2操作。文章包含完整的C++代码实现及调试经验分享。

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题目链接

洛谷P3832

题解

字符串哈希然后丢到hash表里边查询即可
因为\(k \le 50\)，1、2操作就暴力维护一下

经复杂度分析会发现直接这样暴力维护是对的

一开始自然溢出WA了，还以为是哈希冲突，改成双哈希后依旧WA
后来才发现是sb了漏了一句QAQ
不卡自然溢出

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define ULL unsigned long long int
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define res register
using namespace std;
const int maxn = 200005,maxh = 10000005,INF = 1000000000;
const int md = 998244353,P = 19260817;
inline int read(){
    res int out = 0,flag = 1; res char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
inline void write(int x){
    if (x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
struct Hash_Chart{
    ULL num[maxh];
    int head[P + 5],nxt[maxh],cnt,sum[maxh];
    void ins(ULL x){
        int id = x % P;
        if (!head[id]){
            head[id] = ++cnt;
            num[cnt] = x;
            sum[cnt] = 1;
            return;
        }
        for (res int k = head[id]; k; k = nxt[k]){
            if (num[k] == x) {sum[k]++; return;}
            if (!nxt[k]){
                nxt[k] = ++cnt;
                num[cnt] = x;
                sum[cnt] = 1;
                return;
            }
        }
    }
    void del(ULL x){
        int id = x % P;
        for (res int k = head[id]; k; k = nxt[k])
            if (num[k] == x) {sum[k]--; return;}
    }
    int find(ULL x){
        int id = x % P;
        for (res int k = head[id]; k; k = nxt[k]){
            if (num[k] == x) return sum[k];
        }
        return 0;
    }
}hash;
ULL bin[maxn],num[maxn],s1[maxn],s2[maxn];
int lk[maxn],rk[maxn],t1,t2,n,m;
char s[maxh];
inline void solve1(){
    int u = read(),v = read();
    t1 = 0;
    for (res int k = u; k && t1 < 50; k = lk[k]){
        ++t1;
        s1[t1] = s1[t1 - 1] + bin[t1 - 1] * num[k];
    }
    t2 = 0;
    for (res int k = v; k && t2 < 50; k = rk[k]){
        ++t2;
        s2[t2] = s2[t2 - 1] * 7 + num[k];
    }
    for (res int i = 1; i <= t1; i++)
        for (res int j = 1; j <= t2 && i + j <= 50; j++)
            hash.ins(s2[j] + s1[i] * bin[j]);
    rk[u] = v; lk[v] = u;
}
inline void solve2(){
    int u = read(),v = rk[u];
    t1 = 0;
    for (res int k = u; k && t1 < 50; k = lk[k]){
        ++t1;
        s1[t1] = s1[t1 - 1] + bin[t1 - 1] * num[k];
    }
    t2 = 0;
    for (res int k = v; k && t2 < 50; k = rk[k]){
        ++t2;
        s2[t2] = s2[t2 - 1] * 7 + num[k];
    }
    for (res int i = 1; i <= t1; i++)
        for (res int j = 1; j <= t2 && i + j <= 50; j++)
            hash.del(s2[j] + s1[i] * bin[j]);
    rk[u] = lk[v] = 0;
}
inline void solve3(){
    scanf("%s",s + 1); s[0] = '0';
    int len = strlen(s + 1),k = read();
    LL ans = 1; ULL h = 0;
    for (res int i = 1; i <= len; i++){
        h = h * 7 + s[i] - '0';
        if (i >= k){
            h -= bin[k] * (s[i - k] - '0');
            ans = ans * hash.find(h) % md;
        }
    }
    write(ans); putchar('\n');
}
int main(){
    bin[0] = 1; for (res int i = 1; i < maxn; i++) bin[i] = bin[i - 1] * 7;
    n = read(); m = read(); int opt;
    REP(i,n) hash.ins(num[i] = read());
    while (m--){
        opt = read();
        if (opt == 1) solve1();
        else if (opt == 2) solve2();
        else solve3();
    }
    return 0;
}

转载于:https://www.cnblogs.com/Mychael/p/8996326.html